Permutation and combination problems play a major role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to permutation and combination problems. We might have already learned this topic in our school. Even though we have been already taught this topic in our higher classes in school, we need to learn some more short cuts which are being used to solve problems on this topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Permutations:
The ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection are called permutations. |

Combinations:
The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important are called combinations |

Difference between Permutations and Combinations:
Permutations : Selection is made. Beyond selection, order or arrangement is important. Combinations : Selection is made. But order or arrangement is not important |

Formulas:
Permtations : nPr = n!/(n-r)! Combinations : nCr = n!/r!(n-r)! Circular Permutations : = (n-1)! (Both clockwise and anti clockwise rotations are considered. Hint: Every person has the same two neighbors) Circular Permutations : = [(n-1)!]/2 (Either clockwise or anti clockwise rotation is considered.Not both. Hint: No person has the same two neighbors) |

Shortcuts:
1. nPr = n(n-1)(n-2)....to "r" terms Example: 7P3 = 7X6X5 = 210 2. nCr = [n(n-1)(n-2)...to "r" terms]/r! Example: 7P3 = [7X6X5]/[3X2X1] =35 3. nCr = nCn-r (we will use this property only when we want to reduce the value of "r") Example: 25P22 = 25P3 4. nP1 = n 5. nC1 = n 6. nP0 = 1 7. nC0 = 1 8. nPn = n! (No. of permutations of n things taken all at a time) 9. nCn = 1 10. No. of Permutations of n things taken all at a time = (n-1)!.2! (when two things always come together) 11. No. of Permutations of n things taken all at a time = n!-(n-1)!.2! (when two things always not to come together) 12. The value of 0! = 1 12. Fundamental principle of counting: AND ===> Multiplication, OR ===> Addition |

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic permutations and combinations problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve permutation and combination problems. This is the reason for why people must study this topic.

**Here,
we are going to have some permutation and combination problems such that how
shortcuts can be used. You can check your answer online and see step by
step solution.**

1. Compute the sum of all 4 digit numbers which can be formed with the digits 1, 3, 5, 7, if each digit is used only once in each arrangement.

No. of numbers can be formed with the digits 1,3,5,7 = 4! = 4X3X2X1 = 24

No. of digits given = 4 (they are 1,3,5,7)

(No.of numbers formed)/(No.of digits given) = 24/4 = 6

So, each of the given digits will occur six times in each of the place.

Sum of the digits in 1000's place = (1+3+5+7)X 6 = 16X6 = 96

Similar is the case in 100's place, 10's place and 1's place

The sum will be

96X1000 = 96000

96X100 = 9600

96X10 = 960

96X1 = 96

-------------

106656

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Hence, the sum of 4 digits numbers is 106656

No. of digits given = 4 (they are 1,3,5,7)

(No.of numbers formed)/(No.of digits given) = 24/4 = 6

So, each of the given digits will occur six times in each of the place.

Sum of the digits in 1000's place = (1+3+5+7)X 6 = 16X6 = 96

Similar is the case in 100's place, 10's place and 1's place

The sum will be

96X1000 = 96000

96X100 = 9600

96X10 = 960

96X1 = 96

-------------

106656

-------------

Hence, the sum of 4 digits numbers is 106656

2.
The number of 4 digit numbers greater than 5000 can be formed out of the digits 3,4,5,6 and7 (no digit is repeated). The number of such is

Given digits are 3,4,5,6,7 and we form 4 digit numbers greater than 5000.

1000's place --- 3 choices (one of the digits of 5,6,7 is placed)

100's place --- 4 choices (Having 4 out of 5 digits)

10's place --- 3 choices (Having 3 out of 5 digits)

1's place --- 2 choices (Having 2 out of 5 digits)

No. of numbers formed = 3X4X3X2 = 72

Hence, no.of numbers greater than 5000 can be formed is 72

1000's place --- 3 choices (one of the digits of 5,6,7 is placed)

100's place --- 4 choices (Having 4 out of 5 digits)

10's place --- 3 choices (Having 3 out of 5 digits)

1's place --- 2 choices (Having 2 out of 5 digits)

No. of numbers formed = 3X4X3X2 = 72

Hence, no.of numbers greater than 5000 can be formed is 72

3. A examination paper with 10 questions consists of 6 questions in Algebra and 4 questions in geometry. At least one question is to be attempted from each section. In how many ways can this be done?

We have "2" alternatives for each question. That is, either we may attempt or we may not attempt.

Therefore,no. of ways to attempt six questions in Algebra = 2^{6}

(But it includes the way of not attempting all the questions)

So, no. of ways to attempt atleast one question in Algebra = 2^{6}-1

Similarly, no.of ways to attempt atleast one question in Geometry = 2^{4}-1

Total no.ways for both the sections = (2^{6}-1)(2^{4}-1) = 945

Hence, the no. of ways of attempting atleast one question from each section is 945

Therefore,no. of ways to attempt six questions in Algebra = 2

(But it includes the way of not attempting all the questions)

So, no. of ways to attempt atleast one question in Algebra = 2

Similarly, no.of ways to attempt atleast one question in Geometry = 2

Total no.ways for both the sections = (2

Hence, the no. of ways of attempting atleast one question from each section is 945

4.
Find the total number of numbers greater than 2000 can be formed with the digits 1,2,3,4,5, no digit to be repeated in any number.

Given digits are 1,2,3,4, 5.It is not mentioned about how many digits can be used.So,we can form either 4 digit no. or 5 digit no. greater than 2000.

Case1: Forming 4 digit number greater than 2000

1000's place --- 4 choices (one of the digits of 2,3,4,5 is placed)

100's place --- 4 choices (Having 4 out of 5 digits)

10's place --- 3 choices (Having 3 out of 5 digits)

1's place --- 2 choices (Having 2 out of 5 digits)

No. of numbers 4 digit numbers formed = 4X4X3X2 = 96

Case2: Forming 5 digit number greater than 2000

10000's place --- 5 choices (one of the digits of 1,2,3,4,5 is placed)

1000's place --- 4 choices (Having 4 out of 5 digits)

100's place --- 3 choices (Having 3 out of 5 digits)

10's place --- 2 choices (Having 2 out of 5 digits)

1's place --- 1 choice (Having 1 out of 5 digits)

No. of numbers 5 digit numbers formed = 5X4X3X2X1 = 120

Total no. of numbers greater than 2000 formed = 96+120 = 216

Case1: Forming 4 digit number greater than 2000

1000's place --- 4 choices (one of the digits of 2,3,4,5 is placed)

100's place --- 4 choices (Having 4 out of 5 digits)

10's place --- 3 choices (Having 3 out of 5 digits)

1's place --- 2 choices (Having 2 out of 5 digits)

No. of numbers 4 digit numbers formed = 4X4X3X2 = 96

Case2: Forming 5 digit number greater than 2000

10000's place --- 5 choices (one of the digits of 1,2,3,4,5 is placed)

1000's place --- 4 choices (Having 4 out of 5 digits)

100's place --- 3 choices (Having 3 out of 5 digits)

10's place --- 2 choices (Having 2 out of 5 digits)

1's place --- 1 choice (Having 1 out of 5 digits)

No. of numbers 5 digit numbers formed = 5X4X3X2X1 = 120

Total no. of numbers greater than 2000 formed = 96+120 = 216

5. A family of four brothers and three sisters is to be arranged for a photograph in one row. In how many ways can they be seated if no two sisters sit together?

Let us consider the following arrangement.

__ B1__ B2__ B3__ B4__ (B1,B2,B3,B4 --->Brothers)

If we make the sisters to be seated in any of the 3 out of 5 blanks in the above arrangement, no two sisters sit together.

So, sisters can be seated in 5P3 ways

Brothers can be seated in 4!

Thus the total no. ways = 4!X5P3 = (4.3.2.1)X(5.4.3) = 1440

If no two sisters sit together, the no.ways, they can be seated is 1440

__ B1__ B2__ B3__ B4__ (B1,B2,B3,B4 --->Brothers)

If we make the sisters to be seated in any of the 3 out of 5 blanks in the above arrangement, no two sisters sit together.

So, sisters can be seated in 5P3 ways

Brothers can be seated in 4!

Thus the total no. ways = 4!X5P3 = (4.3.2.1)X(5.4.3) = 1440

If no two sisters sit together, the no.ways, they can be seated is 1440

6. A committee of 7 members is to be chosen from 6 artists, 4 singers and 5 writers. In how many ways can this be done if in the committee there must be atleast one member from each group and atleast 3 artists?

For the given condition, possible ways to select members for a committee of 7 members

(3A,3S,1W)==>6C3X4C3X5C1===>20X4X5 = 400

(3A,1S,3W)==>6C3X4C1X3C1===>20X4X10 = 800

(3A,2S,2W)==>6C3X4C2X5C2===>20X6X10 = 1200

(4A,2S,1W)==>6C4X4C2X5C1===>15X6X5 = 450

(4A,1S,2W)==>6C4X4C1X5C2===>15X4X10 = 600

(5A,1S,1W)==>6C5X4C1X5C1===>6X4X5 = 120

Total No.of ways = 400+800+1200+450+600+120 = 3570

Hence, the no. of ways,a committee of 7 members is to be chosen is 3570

(3A,3S,1W)==>6C3X4C3X5C1===>20X4X5 = 400

(3A,1S,3W)==>6C3X4C1X3C1===>20X4X10 = 800

(3A,2S,2W)==>6C3X4C2X5C2===>20X6X10 = 1200

(4A,2S,1W)==>6C4X4C2X5C1===>15X6X5 = 450

(4A,1S,2W)==>6C4X4C1X5C2===>15X4X10 = 600

(5A,1S,1W)==>6C5X4C1X5C1===>6X4X5 = 120

Total No.of ways = 400+800+1200+450+600+120 = 3570

Hence, the no. of ways,a committee of 7 members is to be chosen is 3570

7. The supreme court has given a 6 to 3 decisions upholding a lower court; the number of ways it can give a majority decision reversing the lower court is :

Upholding a lower court means, supporting it for its decision

Reversing a lower court means, opposing it for its decision

In total of 9 cases (6+3=9),it may give 5 or 6 or 7 or 8 or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because majority of 9 is 5 or more.

The possible combinations in which it can give a majority decision reversing the lower court are

5 out of 9 ===> 9C5 = 126

6 out of 9 ===> 9C6 = 84

7 out of 9 ===> 9C7 = 36

8 out of 9 ===> 9C8 = 9

9 out of 9 ===> 9C9 = 1

Total no. of ways = 126+84+36+9+1 = 256

Hence the no. of ways it can a majority of the decision reversing the lower court is 256

Reversing a lower court means, opposing it for its decision

In total of 9 cases (6+3=9),it may give 5 or 6 or 7 or 8 or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because majority of 9 is 5 or more.

The possible combinations in which it can give a majority decision reversing the lower court are

5 out of 9 ===> 9C5 = 126

6 out of 9 ===> 9C6 = 84

7 out of 9 ===> 9C7 = 36

8 out of 9 ===> 9C8 = 9

9 out of 9 ===> 9C9 = 1

Total no. of ways = 126+84+36+9+1 = 256

Hence the no. of ways it can a majority of the decision reversing the lower court is 256

8. Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Number of trials in which the room can be lighted is

Given: 3 bulbs are defective out of 5.There are two bulb points in the dark room

One bulb (or two bulbs) in good condition is enough to light the room.

Since there are two bulb points, we have to select 2 out of 5 bulbs.

No. of ways of selecting 2 bulbs out of 5 = 5P2 = 10

(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)

No. of ways of selecting 2 defective bulbs out of 3 = 3C2 = 3

(It includes selecting only two defective bulbs.So, in these 3 ways, room can not be lighted)

Hence the no. of ways, dark room can be lighted = 10-3 = 7

One bulb (or two bulbs) in good condition is enough to light the room.

Since there are two bulb points, we have to select 2 out of 5 bulbs.

No. of ways of selecting 2 bulbs out of 5 = 5P2 = 10

(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)

No. of ways of selecting 2 defective bulbs out of 3 = 3C2 = 3

(It includes selecting only two defective bulbs.So, in these 3 ways, room can not be lighted)

Hence the no. of ways, dark room can be lighted = 10-3 = 7

9. Find the number of ways of selecting 4 letters from the word EXAMINATION.

There are 11 letters in the word of which A,I,N are repeated twice.

Thus we have 11 letters of 8 different(A,A),(I,I),(N,N),E,X,M,T,O

The group of 4 letters can be selected in any one of the following 4 forms.

(i)2 alike and other 2 alike

(ii)2 alike and other 2 different

(iii)All 4 different

Case(i):If 2 alike and other 2 alike, any 2 of the 3 groups (A,A),(I,I),(N,N) will be selected.

The number of ways = 3C2 = 3C1 = 3

Case(ii):If 2 alike and other 2 different, any one of the three groups (A,A),(I,I),(N,N) and 2 letters from 7 different letters are selected.

[E,X,M,T,O + 2 different letters from(A,A),(I,I),(N,N), because one of the groups is already selected]

The number of ways = 3C1X7C2 = 3X21 = 63

Case(iii): If all four are different, 4 from 8 different letters (A,I,N,E,X,M,T,O) are selected. The number of ways = 8C4 = 70

Hence,the number of ways of selecting 4 letters = 3 + 63 +70 = 136

Thus we have 11 letters of 8 different(A,A),(I,I),(N,N),E,X,M,T,O

The group of 4 letters can be selected in any one of the following 4 forms.

(i)2 alike and other 2 alike

(ii)2 alike and other 2 different

(iii)All 4 different

Case(i):If 2 alike and other 2 alike, any 2 of the 3 groups (A,A),(I,I),(N,N) will be selected.

The number of ways = 3C2 = 3C1 = 3

Case(ii):If 2 alike and other 2 different, any one of the three groups (A,A),(I,I),(N,N) and 2 letters from 7 different letters are selected.

[E,X,M,T,O + 2 different letters from(A,A),(I,I),(N,N), because one of the groups is already selected]

The number of ways = 3C1X7C2 = 3X21 = 63

Case(iii): If all four are different, 4 from 8 different letters (A,I,N,E,X,M,T,O) are selected. The number of ways = 8C4 = 70

Hence,the number of ways of selecting 4 letters = 3 + 63 +70 = 136

10. The letters of the word ZENITH are written in all possible orders. If all the words are written in a dictionary, what is the rank or order of the word ZENITH?

No.of new words formed with the letters of the word ZENITH = 6! = 720

Alphabetical order of the letters of the word ZENITH is E,H,I,N,T,Z

Dictionary gives meanings in the order starting with E,H and so on

Number of words can be formed starting with E,H and so on

E __ __ __ __ __ = 5! = 120 words

H __ __ __ __ __ = 5! = 120 words

I __ __ __ __ __ = 5! = 120 words

N __ __ __ __ __ = 5! = 120 words

T __ __ __ __ __ = 5! = 120 words

Z E H __ __ __ = 3! = 6 words

Z E I __ __ __ = 3! = 6 words

Z E N H __ __ = 2! = 2 words

Z E N I H __ = 1! = 1 word

Z E N I T H = 1! = 1 word

Total Number of words = 120+120+120+120+120+6+6+2+1+1 = 616

Hence, the rank or order of the word ZENITH is 616

Alphabetical order of the letters of the word ZENITH is E,H,I,N,T,Z

Dictionary gives meanings in the order starting with E,H and so on

Number of words can be formed starting with E,H and so on

E __ __ __ __ __ = 5! = 120 words

H __ __ __ __ __ = 5! = 120 words

I __ __ __ __ __ = 5! = 120 words

N __ __ __ __ __ = 5! = 120 words

T __ __ __ __ __ = 5! = 120 words

Z E H __ __ __ = 3! = 6 words

Z E I __ __ __ = 3! = 6 words

Z E N H __ __ = 2! = 2 words

Z E N I H __ = 1! = 1 word

Z E N I T H = 1! = 1 word

Total Number of words = 120+120+120+120+120+6+6+2+1+1 = 616

Hence, the rank or order of the word ZENITH is 616

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